Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n, l[(1 << 20)], r[(1 << 20)], sy, a, b;
int x, y;
inline pair<double, double> getSeg(int ind) {
  return make_pair(x + 1. * (l[ind] - x) * (y - sy) / y,
                   x + 1. * (r[ind] - x) * (y - sy) / y);
}
long long suff[(1 << 20)];
inline long long sum(long long ll, long long rr) {
  if (rr <= ll) {
    return 0;
  }
  return suff[ll] - suff[rr];
  long long ret = 0;
  for (long long i = ll; i < rr; i++) {
    ret += r[i] - l[i];
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout << setprecision(6) << fixed;
  cin >> sy;
  cin >> a >> b;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> l[i] >> r[i];
  }
  for (int i = n - 1; i >= 0; i--) {
    suff[i] = suff[i + 1] + r[i] - l[i];
  }
  long long q;
  cin >> q;
  for (long long i = 0; i < q; i++) {
    cin >> x >> y;
    long long ss = 0, ff = n - 1;
    long long cl = 0, cr = n, mid = (cl + cr) / 2;
    for (; cl + 1 < cr; mid = (cl + cr) / 2) {
      double crr = x + 1. * (r[mid] - x) * (y - sy) / y;
      if (crr >= a) {
        cr = mid;
      } else {
        cl = mid;
      }
    }
    ss = cl;
    cl = 0, cr = n, mid = (cl + cr) / 2;
    for (; cl + 1 < cr; mid = (cl + cr) / 2) {
      double crr = x + 1. * (l[mid] - x) * (y - sy) / y;
      if (crr <= b) {
        cl = mid;
      } else {
        cr = mid;
      }
    }
    ff = cl;
    double ans = 0;
    double factor = (y - sy) * 1. / y;
    long long ps = ss, pf = ff;
    pair<double, double> crr;
    for (; ss <= ff && ss < ps + 3; ss++) {
      crr = getSeg(ss);
      if (crr.second <= a) {
        continue;
      }
      if (crr.first >= b) {
        continue;
      }
      crr.first = max(crr.first, (double)a);
      crr.second = min(crr.second, (double)b);
      ans += crr.second - crr.first;
    }
    for (; ss <= ff && ff > pf - 3; ff--) {
      crr = getSeg(ff);
      if (crr.second <= a) {
        continue;
      }
      if (crr.first >= b) {
        continue;
      }
      crr.first = max(crr.first, (double)a);
      crr.second = min(crr.second, (double)b);
      ans += crr.second - crr.first;
    }
    ans += sum(ss, ff + 1) * factor;
    cout << ans << "\n";
  }
  return 0;
}